555. Split Concatenated Strings 🔒

Description

You are given an array of strings strs. You could concatenate these strings together into a loop, where for each string, you could choose to reverse it or not. Among all the possible loops

Return the lexicographically largest string after cutting the loop, which will make the looped string into a regular one.

Specifically, to find the lexicographically largest string, you need to experience two phases:

1. Concatenate all the strings into a loop, where you can reverse some strings or not and connect them in the same order as given.
2. Cut and make one breakpoint in any place of the loop, which will make the looped string into a regular one starting from the character at the cutpoint.

And your job is to find the lexicographically largest one among all the possible regular strings.

 

Example 1:

Input: strs = ["abc","xyz"]
Output: "zyxcba"
Explanation: You can get the looped string "-abcxyz-", "-abczyx-", "-cbaxyz-", "-cbazyx-", where '-' represents the looped status. 
The answer string came from the fourth looped one, where you could cut from the middle character 'a' and get "zyxcba".

Example 2:

Input: strs = ["abc"]
Output: "cba"

 

Constraints:

• 1 <= strs.length <= 1000
• 1 <= strs[i].length <= 1000
• 1 <= sum(strs[i].length) <= 1000
• strs[i] consists of lowercase English letters.

Solutions

Solution 1: Greedy

We first traverse the string array strs. For each string $s$, if the reversed string $t$ is greater than $s$, we replace $s$ with $t$.

Then we enumerate each position $i$ in the string array strs as a split point, dividing the string array strs into two parts: $strs[i + 1:]$ and $strs[:i]$. We then concatenate these two parts to get a new string $t$. Next, we enumerate each position $j$ in the current string $strs[i]$. The suffix part is $a = strs[i][j:]$, and the prefix part is $b = strs[i][:j]$. We can concatenate $a$, $t$, and $b$ to get a new string $cur$. If $cur$ is greater than the current answer, we update the answer. This considers the case where $strs[i]$ is reversed. We also need to consider the case where $strs[i]$ is not reversed, i.e., concatenate $a$, $t$, and $b$ in reverse order to get a new string $cur$. If $cur$ is greater than the current answer, we update the answer.

Finally, we return the answer.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string array strs.

Python3JavaC++Go

class Solution:
    def splitLoopedString(self, strs: List[str]) -> str:
        strs = [s[::-1] if s[::-1] > s else s for s in strs]
        ans = ''.join(strs)
        for i, s in enumerate(strs):
            t = ''.join(strs[i + 1 :]) + ''.join(strs[:i])
            for j in range(len(s)):
                a = s[j:]
                b = s[:j]
                ans = max(ans, a + t + b)
                ans = max(ans, b[::-1] + t + a[::-1])
        return ans

class Solution {
    public String splitLoopedString(String[] strs) {
        int n = strs.length;
        for (int i = 0; i < n; ++i) {
            String s = strs[i];
            String t = new StringBuilder(s).reverse().toString();
            if (s.compareTo(t) < 0) {
                strs[i] = t;
            }
        }
        String ans = "";
        for (int i = 0; i < n; ++i) {
            String s = strs[i];
            StringBuilder sb = new StringBuilder();
            for (int j = i + 1; j < n; ++j) {
                sb.append(strs[j]);
            }
            for (int j = 0; j < i; ++j) {
                sb.append(strs[j]);
            }
            String t = sb.toString();
            for (int j = 0; j < s.length(); ++j) {
                String a = s.substring(j);
                String b = s.substring(0, j);
                String cur = a + t + b;
                if (ans.compareTo(cur) < 0) {
                    ans = cur;
                }
                cur = new StringBuilder(b)
                          .reverse()
                          .append(t)
                          .append(new StringBuilder(a).reverse().toString())
                          .toString();
                if (ans.compareTo(cur) < 0) {
                    ans = cur;
                }
            }
        }
        return ans;
    }
}

class Solution {
public:
    string splitLoopedString(vector<string>& strs) {
        for (auto& s : strs) {
            string t{s.rbegin(), s.rend()};
            s = max(s, t);
        }
        int n = strs.size();
        string ans = "";
        for (int i = 0; i < strs.size(); ++i) {
            auto& s = strs[i];
            string t;
            for (int j = i + 1; j < n; ++j) {
                t += strs[j];
            }
            for (int j = 0; j < i; ++j) {
                t += strs[j];
            }
            for (int j = 0; j < s.size(); ++j) {
                auto a = s.substr(j);
                auto b = s.substr(0, j);
                auto cur = a + t + b;
                if (ans < cur) {
                    ans = cur;
                }
                reverse(a.begin(), a.end());
                reverse(b.begin(), b.end());
                cur = b + t + a;
                if (ans < cur) {
                    ans = cur;
                }
            }
        }
        return ans;
    }
};

func splitLoopedString(strs []string) (ans string) {
    for i, s := range strs {
        t := reverse(s)
        if s < t {
            strs[i] = t
        }
    }
    for i, s := range strs {
        sb := &strings.Builder{}
        for _, w := range strs[i+1:] {
            sb.WriteString(w)
        }
        for _, w := range strs[:i] {
            sb.WriteString(w)
        }
        t := sb.String()
        for j := 0; j < len(s); j++ {
            a, b := s[j:], s[0:j]
            cur := a + t + b
            if ans < cur {
                ans = cur
            }
            cur = reverse(b) + t + reverse(a)
            if ans < cur {
                ans = cur
            }
        }
    }
    return ans
}

func reverse(s string) string {
    t := []byte(s)
    for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
        t[i], t[j] = t[j], t[i]
    }
    return string(t)
}

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